的reinterpret_cast<> [英] reinterpret_cast<>

问题描述

在SunOS 5.9上编写了这段代码，编译器g ++ 2.95.3


试图查看int或short的字节顺序int转换为

char *。不行。 char * cpt似乎没有初始化!!。

为什么会这样？

int main（）

{

int x = 0x01020304;

int * ipt =& x;

cout<< ipt：<< hex<< ipt<<结束// ok ---

char * cpt = reinterpret_cast< char *>（ipt）;

cout<< cpt： << hex<< cpt<< endl; //没有cpt的输出!!!

short int * spt = reinterpret_cast< short int *> （ipt）;

cout<< spt： << hex<< spt<< endl; // ok ---


}


问候，

阿曼。

-

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解决方案

" Aman" <上午** @ techie.com>在留言中写道

新闻：40 ************************************ @mygate .mailgate.org ...

在SunOS 5.9上编写了这段代码，编译器g ++ 2.95.3
试图查看字节转换为
char *的int或short int的顺序。不行。 char * cpt似乎没有被初始化!!
为什么会这样？


它是。

int main（）
{x /> int x = 0x01020304;
int * ipt =& x;
cout<< ipt：<< hex<< ipt<<结束// ok ---
char * cpt = reinterpret_cast< char *>（ipt）;
cout<< cpt： << hex<< cpt<< endl; //没有输出为cpt !!!

类型''char *''的流插入器（<<）解释了参数

作为指向C风格字符串的指针（零终止字符数组）。

例如

char * p =" hello" ;;

cout<< p << \\\
; / *打印你好 * /


有两件事正在发生：


- 系统上的sizeof（int）是四个字节，在这种情况下没有

的值为零，产生''undefined behavior''，因为

cout<<将运行你的数组结束。


-sizeof（int）超过四个字节，在这种情况下，其中一些

的值为零，并且看起来在你的机器上，在任何非零字节之前存储了零

，在这种情况下，

cout<< cpt在任何输出之前终止。


尝试：


for（size_t i = 0; i< sizeof x; ++ i）

cout<< hex<< cpt [i];

cout<< ''\\ n';;

-Mike

" Mike Wahler" < MK ****** @ mkwahler.net>在消息中写道

news：x4 ***************** @ newsread1.news.pas.earthl ink.net ...

尝试：

for（size_t i = 0; i< sizeof x; ++ i）
cout<< hex<< cpt [i];

我们需要转换为int来获取

的文本表示字节'的值：


cout<< hex<< static_cast< int>（cpt [i]）;


对不起的疏忽。


-Mike

>试图通过转换为

char *来查看int或short int的字节顺序

为什么不使用联合？

Hi,
wrote this piece of code on SunOS 5.9 , compiler g++ 2.95.3

trying to see the byte order of an int or short int by converting to
char* . doesn''t work . the char* cpt doesn''t seem to be initialized !!.
why would that be ?
int main()
{
int x=0x01020304 ;
int* ipt = &x ;
cout << "ipt : "<< hex << ipt << endl ; // ok ---
char* cpt = reinterpret_cast<char*>(ipt) ;
cout << "cpt : " << hex << cpt <<endl ; // NO output for cpt !!!
short int* spt = reinterpret_cast<short int*> (ipt) ;
cout << "spt : " << hex << spt <<endl ; // ok ---

}

regards,
Aman.
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解决方案

"Aman" <am**@techie.com> wrote in message
news:40************************************@mygate .mailgate.org...

Hi,
wrote this piece of code on SunOS 5.9 , compiler g++ 2.95.3

trying to see the byte order of an int or short int by converting to
char* . doesn''t work . the char* cpt doesn''t seem to be initialized !!.
why would that be ?
It is.


int main()
{
int x=0x01020304 ;
int* ipt = &x ;
cout << "ipt : "<< hex << ipt << endl ; // ok ---
char* cpt = reinterpret_cast<char*>(ipt) ;
cout << "cpt : " << hex << cpt <<endl ; // NO output for cpt !!!

The stream inserter (<<) for type ''char*'' interprets the argument
as a pointer to a ''C-style'' string (zero terminated array of char).
e.g.
char *p = "hello";
cout << p << ''\n''; /* prints "hello" */

On of two things is happening:

- sizeof(int) on your system is four bytes, in which case none
of them has a value of zero, producing ''undefined behavior'', since
cout<< will run off the end of your array.

-sizeof(int) is more than four bytes, in which case some of them
have zero value, and it appears that on your machine, a zero
is stored before any of the nonzero bytes, in which case
cout << cpt terminates before any output.

Try:

for(size_t i = 0; i < sizeof x; ++i)
cout << hex << cpt[i];
cout << ''\n'';
-Mike

"Mike Wahler" <mk******@mkwahler.net> wrote in message
news:x4*****************@newsread1.news.pas.earthl ink.net...

Try:

for(size_t i = 0; i < sizeof x; ++i)
cout << hex << cpt[i];

We need to cast to int to get the textual representation of
the byte''s value:

cout << hex << static_cast<int>(cpt[i]);

Sorry for the oversight.

-Mike

> trying to see the byte order of an int or short int by converting to

char*

Why not use a union?

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